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Uniform Acceleration

Uniform Acceleration. Motion in One Dimension. Sects. 5-7: Outline. 1. Kinematics of one dimensional motion with uniform acceleration:. 2. Special case: Freely falling objects  a = -g = - 9.8 m/s 2. 3. Special case: a = 0  motion with uniform velocity v = constant. Uniform Acceleration.

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Uniform Acceleration

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  1. Uniform Acceleration

  2. Motion in One Dimension Sects. 5-7: Outline 1. Kinematics of one dimensional motion with uniform acceleration: 2. Special case: Freely falling objects  a = -g = - 9.8 m/s2 3. Special case: a = 0 motion with uniform velocityv = constant

  3. Uniform Acceleration • The derivation is in the text twice, once using algebra & once using calculus. Read on your own! • Notation(drop subscripts x on v & a, for motion along x only!) t  0 = time when the problem begins xiinitial position (at t = 0, often xi = 0) viinitial velocity (at t1= 0) t  time when we wish to know other quantities xfposition at time t vfvelocity at time t a  acceleration = constant (Average & instantaneous accelerations are equal)

  4. Uniform Acceleration Equations • Results (one dimensional motiononly!): vf = vi + at (1) xf = xi + vi t + (½)a t2 (2) (vf)2 = (vi)2 + 2a(x - x0) (3) vavg = (½)(vf + vi) (4) NOT VALIDUNLESS a = CONSTANT!!! Often xi = 0. Sometimes vi = 0

  5. Physics and Equations • IMPORTANT!!! • Even though these equations & their applications are important, Physics is nota collection of formulas to memorize & blindly apply! • Physics is a set of PHYSICAL PRINCIPLES. • Blindly searching for the “equation which will work for this problem” is DANGEROUS!!!! • On exams, you get to have an 3´´ 5´´ index card with anything written on it (both sides) you wish. On quizzes, I will give you relevant formulas.

  6. Example: Acceleration of Car Known: x0 = 0, x = 30 m, v0 = 0, a = 2.0 m/s2 Wanted: t Use: x = (½)a t2  t = (2x/a)½ = 5.48 s

  7. Example: Estimate Breaking Distances   a = - 6.0 m/s2 v decreases from 14 m/s to zero x0 = 7 m, v0 = 14 m/s, v = 0 v2 = (v0)2 + 2a(x – x0)  x = x0 + [v2 - (v0)2]/(2a) = 7 m + 16 m = 23 m v = v0 = constant = 14 m/s t = 0.50 s a = 0 x = v0t = 7 m

  8. Example: Fastball Known: x0 = 0, x = 3.5 m, v0 = 0, v = 44 m/s Wanted: a Use: v2 = (v0)2 + 2a (x - x0)  a = (½)[v2 - (v0)2]/(x - x0) = 280 m/s2 !

  9. Example 2.7: Carrier Landing, p. 35 Problem: A jet lands on an aircraft carrier at 140 mi/h (63 m/s). a) Calculate the acceleration (assumed constant) if it stops in t = 2.0 s due to the arresting cable that snags the airplane & stops it. b) If the plane touches down at position xi = 0, calculate it’s final position.

  10. Example 2.8, Watch Out for the Speed Limit! p. 39 Problem: A car traveling at a constant velocity of magnitude 41.4 m/s passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch it, accelerating at a constant rate of 3.90 m/s2. How long does it take her to overtake the speeding car?

  11. Freely Falling Objects

  12. Freely Falling Objects • Important & common special case of uniformly accelerated motion: “FREE FALL” Objects falling in Earth’s gravity. Neglect air resistance. Use one dimensional uniform acceleration equations (with some changes in notation, as we will see)

  13. A COMMON MISCONCEPTION!

  14. Experiment finds that the acceleration of falling objects(neglecting air resistance)is always(approximately)the same, no matter how light or heavy the object. • Acceleration due to gravity, a  g g = 9.8 m/s2(approximately!)

  15. Acceleration of falling objects is always the same, no matter how light or heavy. • Acceleration due to gravity, g = 9.8 m/s2 • First proven by Galileo Galilei Legend: Dropped objects off of the leaning tower of Pisa.

  16. Acceleration due to gravity g = 9.8 m/s2(approximately) • Depends on location on Earth, latitude, & altitude:

  17. Note: My treatment is slightly different than the book’s, but it is, of course, equivalent! • To treat motion of falling objects, use the same equations we already have, but change notation slightly: Replace a by g = 9.8 m/s2 • But in the equations it could have a + or a - sign in front of it! Discuss this next! • Usually, we consider vertical motion to be in the y direction, so replace xf by yf and xi by yi(often y0i = 0)

  18. NOTE!!! Whenever I write the symbol g, itALWAYSmeans thePOSITIVEnumerical value 9.8 m/s2! ItNEVERis negative!!! The sign (+ or -) of the gravitational acceleration is taken into account in the equations we now discuss!

  19. Sign of g in 1d Equations • Magnitude (size) of g = 9.8 m/s2(POSITIVE!) • But, acceleration is avector(1 dimensional), with 2 possible directions. • Call these + and -. • However, which way is + and which way is - is ARBITRARY & UP TO US! • May seem “natural” for “up” to be + y and “down” to be - y, but we could also choose (we sometimes will!) “down” to be + y and “up” to be - y • So, in equations g could have a + or a - sign in front of it, depending on our choice!

  20. Directions of Velocity & Acceleration • Objects in free fall ALWAYShave downward acceleration. • Still use the same equations for objects thrown upward with some initial velocity vi • An object goes up until it stops at some point & then it falls back down. Acceleration is alwaysg in the downward direction. For the first half of flight, the velocity is UPWARD.  For the first part of the flight, velocity & acceleration are in opposite directions!

  21. VELOCITY & ACCELERATION ARENOT NECESSARILYIN THE SAME DIRECTION!

  22. Equations for Bodies in Free Fall • Written taking “up” as+ y! vf = vi - g t (1) yf = yi + vit – (½)gt2 (2) (vf)2 = (vi)2 - 2g(yf - yi) (3) vavg = (½)(vf + vi) (4) g = 9.8 m/s2 Often, yi = 0. Sometimes vi = 0

  23. Equations for Bodies in Free Fall • Written taking “down” as+ y! vf = vi + g t (1) yf = yi + vit + (½)gt2 (2) (vf)2 = (vi)2 + 2g(yf - yi) (3) vavg = (½)(vf + vi) (4) g = 9.8 m/s2 Often, yi = 0. Sometimes vi = 0

  24. Example v1 = (9.8)(1) = 9.8 m/s Note:y is positive DOWNWARD! v = at y = (½) at2 a = g = 9.8 m/s2 v2 = (9.8)(2) = 19.6 m/s v3 = (9.8)(3) = 29.4 m/s

  25. Example

  26. Example 2.12: Not a bad throw for a rookie! p. 38 Problem: A stone is thrown at point (A) from the top of a building with an initial velocity of vi = 19.2 m/s straight upward. The building is H = 49.8 m high, and the stone just misses the edge of the roof on its way down, as in the figure. Answer these questions: a) Calculate the time at which the stone reaches its maximum height. b) Calculate the maximum height of the stone above the rooftop. c) Calculate the time at which the stone returns to the level of the thrower d) Calculate the velocity of the stone at this instant. e) Calculate the velocity & position of the stone at time t = 5 s.

  27. Motion in One Dimension Sects. 5-7: Summary 1. Kinematics of one dimensional motion with uniform acceleration: 2. Special case: Freely falling objects  a = -g = - 9.8 m/s2 3. Special case: a = 0 motion with uniform velocityv = constant

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